Correct Answer: D. (-\(\frac{3}{8}\), \(\frac{3}{2}\))

Correct Answer: D. 16√3cm²

Explanation

|AB| = 4cm

|BC| = 8cm

In right-angled BAC;

8\(^2\) = 4\(^2\) + |AC|\(^2\)

|AC|\(^2\) = 8\(^2\) - 4\(^2\)

|AC\\(^2\) = 64 - 16 → 48

|AC| = \(\sqrt{48}\)cm → \(4\sqrt{3}cm\)

The area of rectangle = L x B

= |AB| x |AC|

= \((4 \times 4\sqrt{3}cm^2\)

=16\sqrt{3}cm^2\)

Correct Answer: A. \(12x^2-5x-2=0\)

Explanation

x\(^2\) - (sum of given roots)x + (product of given roots ) = 0

x\(^2\) - (\(\frac{2}{3} + \frac{-1}{4}\))x + (\(\frac{2}{3} * \frac{-1}{4}\)) = 0

x\(^2\) - (\(\frac{8 - 3}{12}\))x + (\(\frac{- 2}{12}\))

x\(^2\) - \(\frac{5}{12}\)x + \(\frac{- 2}{12}\)

multiply through by the LCM 12

12 * x\(^2\) - 12 * \(\frac{5}{12}\)x + 12 * \(\frac{- 2}{12}\)

12x\(^2\) - 5x - 2 = 0

Explanation

(a) \(\frac{x^{2} - 4}{x^{2} + x} \div \frac{x^{2} - 4x + 4}{x + 1}\)

= \(\frac{x^{2} - 4}{x^{2} + x} \times \frac{x + 1}{x^{2} - 4x + 4}\)

= \(\frac{(x - 2)(x + 2)}{x(x + 1)} \times \frac{x + 1}{x^{2} - 2x - 2x + 4}\) (Using difference of two squares)

= \(\frac{(x - 2)(x + 2)}{x} \times \frac{1}{(x - 2)^{2}}\)

= \(\frac{(x + 2)}{x(x - 2)}.

(b) Roots of \(ax^{2} +bx + c = mx + k\)

For graph \(y = ax^{2} + bx + c\),

\(x = -1\) and \(x = 2\)

\(x + 1 = 0\) and \(x - 2 = 0\)

\((x + 1)(x - 2) = 0\)

\(x^{2} - 2x + x - 2 = 0\)

\(x^{2} - x - 2 = 0 ..... (i)\)

\(-x^{2} + x + 2 = 0 ...... (ii)\)

For graph \(y = mx + k\)

\(m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{1 - (-1)}{1.3 - (-1.3)}\)

\(m = \frac{2}{2.6} = 0.769\)

\(k = 0.25\)

\(x^{2} - x - 2 = 0.769x + 0.25\)

\(x^{2} - x - 0.769x - 2 - 0.25 = 0\)

\(x^{2} - 1.769x - 2.25 = 0\)

Using formula method, \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)

\(x = \frac{-(-1.769) \pm \sqrt{(-1.769)^{2} - 4(1)(-2.25)}}{2(1)}\)

\(x = \frac{1.769 \pm \sqrt{3.1294 + 9}}{2}\)

\(x = \frac{1.769 \pm 3.483}{2}\)

\(x = \frac{1.769 + 3.483}{2}\) or \(x = \frac{1.769 - 3.483}{2}\)

\(x = \frac{5.252}{2} \approxeq 2.626\) or \(x = \frac{-1.714}{2} \approxeq -0.857\)

(ii) The coordinates of points L, M and N are (-1, 0), (0, 2) and (2, 0) respectively.

Using coordinate for L:

\(0 = a(-1^{2}) + b(-1) + c \implies 0 = a - b + c ..... (1)\)

Using M :

\(2 = a(0^{2}) + b(0) + c \implies c = 2 ...... (2)\)

Using N :

\(0 = a(2^{2}) + b(2) + c \implies 0 = 4a + 2b + c .... (3)\)

From (2), c = 2

\(\therefore (1) : a - b + 2 = 0 \implies a - b = -2 .... (1a)\)

\(\therefore (3) : 4a + 2b + 2 = 0 \implies 4a + 2b = -2 ..... (3a)\)

Solving equations (1a) and (3a) simultaneously, we get

\(a = -1 , b = 1 , c = 2\)

\(\therefore\) Equation of the curve : \(-x^{2} + x + 2 = 0\)

(iii) Line of symmetry = \(x = \frac{-b}{2a}\)

\(y = -x^{2} + x + 2 \)

Line of symmetry = \(\frac{-1}{2(-1)}\)

\(x = \frac{1}{2}\)

Correct Answer: A. 113\(cm^3\)

Explanation

Surface area of a sphere = \(4 \pi r^2\)

\(4 \pi r^2\) = \(\frac{792}{7}cm^2\)

4 x \(\frac{22}{7}\) x \(r^2\) = \(\frac{792}{7}\)

\(r^2\) = \(\frac{792}{7}\) x \(\frac{7}{4 \times 22}\)

= 9

r = \(\sqrt{9}\)

= 3cm

Hence, volume of sphere

= \(\frac{4}{3} \pi r^3\)

= \(\frac{4}{3} \times \frac{22}{7} \times 3 \times 3 \times 3 \)

= \(\frac{4 \times 22 \times 9}{7}\)

\(\approx\) = 113.143

= 113\(cm^3\) (to the nearest whole number)

Correct Answer: C. cos 285°

Explanation

Spearman's rank correlation coefficient = \(1 - \frac{6 \sum d^{2}}{n(n^{2} - 1)}\)

= \(1 - \frac{6(10)}{5(5^{2} - 1)}\)

= \(1 - \frac{60}{120}\)

= \(0.5\)

Correct Answer: B. x ≤ -3 or x ≥ 4

Correct Answer: A. 40°

Correct Answer: D. 7/3